Unsplitting sequence of vector bundles

Let $$V$$ be a $$n$$-dimensional complex vector space. Using Grothendieck's notation, we define the Grassmannian $$G(k,V)$$ as the space of $$k$$-quotients of $$V$$ or, equivalently, as $$G(k,V)=\{ \mathbb P W \subset \mathbb P V: \dim W=k\}.$$ On $$G(k,V)$$ we have the tautological bundle sequence: $$0 \to \mathcal S^\vee \to V \otimes \mathcal O_{G(k,V)} \to \mathcal Q \to 0.$$ We suppose that the following short exact sequence holds: $$0 \to \mathcal Q \otimes \mathcal S^\vee \to N \to \wedge^2 \mathcal Q \to 0,$$ where $$N$$ is another vector bundle on $$G(k,V)$$. I know that in the case $$k=2$$ $$Ext^1(\wedge^2 \mathcal Q, \mathcal Q \otimes \mathcal S^\vee)=H^1(G(k,V), \mathcal Q \otimes \mathcal S^\vee \otimes (\wedge^2 \mathcal Q)^\vee)$$ become, using $$\wedge^2 \mathcal Q=\mathcal O(1)$$ and $$\mathcal Q(-1)=\mathcal Q^\vee$$, $$Ext^1(\mathcal O(1), \mathcal Q \otimes \mathcal S^\vee)=H^1(G(k,V),\mathcal Q^\vee \otimes \mathcal S^\vee)=H^1(G(k,V), \Omega^1)=\mathbb C,$$ hence the short exact sequence do not split. I expect that this happens also in the general case, but I cannot prove it. How can I proceed?

• Which short exact sequence do you want to show does not split? You did not define the vector bundle $N$ (as far as I can tell), so how do we know $N$ is not $\mathcal{Q}\otimes \mathcal{S}^\vee \oplus \wedge^2\mathcal{Q}$? Commented Sep 1, 2021 at 15:25
• You right! $N$ is the normal bundle of the Grassmannian $G(k,V)$ into an orthogonal Grassmannian $OG(k,V \oplus V')$. I don't know an explicit description of $N$, I only know that it is the middle term of this short exact sequence Commented Sep 2, 2021 at 7:48
• It seems likely that your sequence is given by taking $N=\wedge^2 (V\otimes \mathcal{O})/\wedge^2(\mathcal{S}^\vee)$. Can you prove that? This bundle would give rise to such an exact sequence. Commented Sep 2, 2021 at 12:18
• I don't understand why it should be such a bundle, even on the fiber level. By the way, I have to understand what is $Ext^1$, because it parametrize the extensions: if it is zero, then $N$ has to the direct sum. My idea is that there is only one extension as in the case $k=2$, that is an Atiyah extension Commented Sep 2, 2021 at 14:39

$$\def\CC{\mathbb{C}}$$A splitting would be a global map $$f : G(k,n) \times \CC^n \to \CC^n$$ such that $$f(L,v) \in L$$ for all $$L \in G(k,n)$$ and $$v \in \CC^n$$. But, since $$G(k,n)$$ is projective and connected and the target $$\CC^n$$ is affine, any map $$f : G(k,n) \times \CC^n \to \CC^n$$ is constant on each $$G(k,n) \times \{ v \}$$. So we would have a map $$f : \CC^n \to \CC^n$$ such that $$f(v) \in L$$ for each $$v \in \CC^n$$ and each $$L$$ in $$G(k,n)$$. The only such map is the zero map.